Answer :
Answer:
a) 0.3520 = 35.20% probability that a hotel room costs $225 or more per night.
b) 0.1038 = 10.38% probability that a hotel room costs less than $140 per night.
c) 0.4878 = 48.78% probability that a hotel room costs between $200 and $300 per night.
d) $251.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 204, \sigma = 55[/tex]
a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?
This probability is 1 subtracted by the pvalue of Z when X = 225. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{225 - 204}{55}[/tex]
[tex]Z = 0.38[/tex]
[tex]Z = 0.38[/tex] has a pvalue of 0.6480.
1 - 0.6480 = 0.3520
0.3520 = 35.20% probability that a hotel room costs $225 or more per night.
b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)?
This probability is the pvalue of Z when X = 140. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 204}{55}[/tex]
[tex]Z = -1.16[/tex]
[tex]Z = -1.16[/tex] has a pvalue of 0.1038.
0.1038 = 10.38% probability that a hotel room costs less than $140 per night.
c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)?
This probability is the pvalue of Z when X = 300 subtracted by the pvalue of Z when X = 200. So
X = 300
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{300 - 204}{55}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a pvalue of 0.9599.
X = 200
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 204}{55}[/tex]
[tex]Z = -0.07[/tex]
[tex]Z = -0.07[/tex] has a pvalue of 0.4721.
0.9599 - 0.4721 = 0.4878
0.4878 = 48.78% probability that a hotel room costs between $200 and $300 per night.
d. What is the cost of the 20% most expensive hotel rooms in New York City?
This is X when Z has a pvalue of 1-0.2 = 0.8. So X when Z = 0.84
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 204}{55}[/tex]
[tex]X - 204 = 0.84*55[/tex]
[tex]X = 250.2[/tex]
Rounding to the next dollar.
$251.